# Divergence of a Curl - Then Integrate By Parts!

1. Oct 21, 2012

### Chingon

1. The problem statement, all variables and given/known data

Prove this by integration by parts. A(r) and B(r) vanish at infinity.

2. Relevant equations
I'm getting stuck while trying to integrate by parts - I end up with partial derivatives and dV, which is dxdydz?

3. The attempt at a solution

I can break things down to Cartesian components, but integrating by parts is where I get stuck.

Essentially, I'm simplifying by stating the identity that
BdotCurlA - AdotCurlB = Div(A×B) = 0 in this case (Subtract right hand side from left and combine under one ∫dV)

The components look like this:
∂x(AyBz−AzBy)=(∂xAy)Bz+Ay(∂xBz)−(∂xAz)By−Az(∂xBy)
plus the y,z terms as well.

How would one integrate ∫[(∂xAy)Bz]dV by parts? I have Ay and Bz which are potentially functions of x,y,z and a partial wrt x, and dxdydz...

Thanks!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 21, 2012

### dextercioby

Do you know cartesian tensors ? If so, the problem is trivial.

$$\iiint A_k \epsilon_{klm} \partial_{l}B_{m} dV = ...$$

3. Oct 21, 2012

### lurflurf

Integrations by parts are equivalent to product rules.
use
$$\nabla \cdot (\mathbf{A} \times \mathbf{B}) = \mathbf{B} \cdot (\nabla \times \mathbf{A}) - \mathbf{A} \cdot (\nabla \times \mathbf{B})$$
and the divergence theorem to reduces the problem to showing a surface integral at infinity is zero.

4. Oct 22, 2012

### Chingon

No, we haven't studied those so I don't believe that's a path I can take.

I have used that identity to reduce the problem to ∫Div(A×B)=0

This reduces to:
∫[∂xAy*Bz-∂xAz*By+∂yAz*By-∂yAx*BZ+∂ZAx*By-∂zAy*Bx]dV

5. Oct 22, 2012

### Chingon

So I guess I'm wondering how to handle the partial derivatives inside the integrand. Take the first term for example:
∫∂/∂x(AyBz)dxdydz

Do the ∂x and dx cancel out?

6. Oct 22, 2012

### dextercioby

No, no, no. Do you know about the Gauß-Ostrogradskii theorem ? It turns a volume integral into a surface integral.

Last edited: Oct 22, 2012
7. Oct 22, 2012

### Chingon

Yes, but I was focused on the brute force of integration by parts method as opposed to using this theorem. I think I'm just going to go the easy route and use this method - hopefully I won't lose too many points!

8. Oct 22, 2012

### dextercioby

You are doing vector integrals, you should use the vectors not their components, and you can use the Gauß-Ostrogradskii theorem in case of vectors.

<In vector calculus, the divergence theorem, also known as Gauss's theorem or Ostrogradsky's theorem,[1] is a result that relates the flow (that is, flux) of a vector field through a surface to the behavior of the vector field inside the surface.>

No reason to lose points.